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Does The Decomp Temp Drop Under Vacuum ?


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#1 Guest_Shanker Nair_*

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Posted 14 September 2004 - 05:11 AM

Suppose a substacne cracks at 250 C at atm pressure.

Now under vacuum does the decomposition / cracking temperature of this substance drop ???


#2 siretb

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Posted 17 September 2004 - 03:24 AM

probably not. decomposition is purely a thermal process, that I would expect be governed by Arrhenius law (kinetics)

#3 djack77494

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Posted 18 September 2004 - 07:46 PM

Shanker,
You may possibly be confused by having heard how vacuum conditions are often employed when there is the possibility of thermal decomposition. Vacuum is employed so that the heat sensitive material may be vaporized at lower temperatures, thereby minimizing the risks of thermal decomposition. If the temperature was not decreased, the risk of thermal decomposition would be the same or even higher, since vacuum conditions would probably promote decomposition.

Doug Jackson

#4 Guest_Shanker Nair_*

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Posted 19 September 2004 - 09:50 AM

Thank you for your replies !

well, jack, I'am really sorry , but I'am a little confused! What I decipher out of what you said, is that vacuum is used so as to avoid a substance decomposing due to heat (by reducing its temperature ) Is that it ????

And later you say again vacuum conditions promote decomposition. Now for instance, in my case , there is this HIGHLY HIGHLY EXPLOSIVE material, which has to be cracked at abt 250 C , into a product that has a bp of 298C.

Now , i was wondering , by maintaining vacuum conditions, if the decomposition temperature could be brought down to say 120 C or something like that, it would be much safer to operate !

#5 mbeychok

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Posted 20 September 2004 - 04:42 PM

Shanker Nair:

Let me try to explain what DJack said:

(1) To crack a liquid substance, in general you must first vaporize the substance.

(2) Let us assume that your substance is a liquid with an atmospheric boiling point of 400°C. Thus, to vaporize the material under atmospheric pressure, you would have to heat it to 400°C ... which is above your decomposition temperature of 250°C, and that is unacceptable to you.

(3) However, the substance can be vaporized at a lower temperature by vaporizing it under vacuum conditions ... because the boiling point of any liquid is lowered as you reduce the pressure. Thus, you could vaporize the material at some vacuum where its boiling point is less than 250°C.

(4)Thus, lowering the pressure does not change the decompositon temperature but it allows you to vaporize the liquid at a temperature below the decompositon temperature ... which is what you want to do.

Now, all of the above is based on your substance being a liquid and that it needs to be vaporized before it will crack.

#6 Guest_Shanker Nair_*

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Posted 21 September 2004 - 12:04 PM

Oh well, thankyou, SO MUCH for your time. I have a clear picture now ! THANKYOU !

Anyway, I just wanetd to elaborate ! What we have is a 40% solution (consisting of trimer(solute) and diethyl malonate(DEM)(solvent)) that decomposes at 250 C. DEM has a BP of 198 C. Our interest is to decompose the trimer and thus obtain the desired product (btw, without adding DEM, the trimer is a crystallien mass and decomposes at 130 C, (and is highly highly explosive).

So, here what we need is DECOMPOSITION, we are not trying to avoid decomposition. My colleagues (seniors) in the lab feel applying vacuum would help to decompose this substance somewhere at abt 150 C. So that DEM is separated at 80 C(from Perry's under vacuum) , and also the cracked/decomposed product is obtained!

Anyway, thans once again smile.gif

#7 Guest_ronald reagan_*

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Posted 24 August 2006 - 11:37 PM

QUOTE (djack77494 @ Sep 18 2004, 07:46 PM) <{POST_SNAPBACK}>
Shanker,
You may possibly be confused by having heard how vacuum conditions are often employed when there is the possibility of thermal decomposition. Vacuum is employed so that the heat sensitive material may be vaporized at lower temperatures, thereby minimizing the risks of thermal decomposition. If the temperature was not decreased, the risk of thermal decomposition would be the same or even higher, since vacuum conditions would probably promote decomposition.

Doug Jackson


#8 Guest_ronald reagan_*

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Posted 24 August 2006 - 11:42 PM

QUOTE (djack77494 @ Sep 18 2004, 07:46 PM) <{POST_SNAPBACK}>
Shanker,
You may possibly be confused by having heard how vacuum conditions are often employed when there is the possibility of thermal decomposition. Vacuum is employed so that the heat sensitive material may be vaporized at lower temperatures, thereby minimizing the risks of thermal decomposition. If the temperature was not decreased, the risk of thermal decomposition would be the same or even higher, since vacuum conditions would probably promote decomposition.

Doug Jackson

For a case of having a vegetable inside a vacuum tank. Is the time span of decomposition of the vegetable increases or decreases? [quote]

#9 Guest_Guest_*

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Posted 27 August 2006 - 03:01 AM

QUOTE (Shanker Nair @ Sep 21 2004, 12:04 PM) <{POST_SNAPBACK}>
Oh well, thankyou, SO MUCH for your time. I have a clear picture now ! THANKYOU !

Anyway, I just wanetd to elaborate ! What we have is a 40% solution (consisting of trimer(solute) and diethyl malonate(DEM)(solvent)) that decomposes at 250 C. DEM has a BP of 198 C. Our interest is to decompose the trimer and thus obtain the desired product (btw, without adding DEM, the trimer is a crystallien mass and decomposes at 130 C, (and is highly highly explosive).

So, here what we need is DECOMPOSITION, we are not trying to avoid decomposition. My colleagues (seniors) in the lab feel applying vacuum would help to decompose this substance somewhere at abt 150 C. So that DEM is separated at 80 C(from Perry's under vacuum) , and also the cracked/decomposed product is obtained!

Anyway, thans once again smile.gif

Shanker Nair:
The above context quoted meminds me of the following:

(1) The monomer of DEM (also dimer ) and trimer are in an equilibrium state
(M, D, T are monomer,dimer and trimer respectively):
M <--> D <--> T
(2) The "decomposition" of trimer means the reaction:
T--> D -->M in the liquid phase
(3) But the "DECOMPOSITION" of trimer in the solid(or crystalline) or vapor phase means the reaction:
T--> CO2, H2O, ...etc.
(4) There would be a threshold concentration of trimer in the liquid phase for "DECOMPOSITION" with respect to each temperature of trimer solution. (Below 130℃, the trimer will not "DECOMPOSE")
(5) There would be a threshold temperature of solution in the liquid phase for "DECOMPOSITION" with respect to each concentration of trimer (The trimer crystal, 100% trimer, DECOMPOSEs at 130 ℃)
(6) And, you are going to get only 'M' thru evaproation (or distillation) under vacuum from the mixture of M, D, and T. During the process, reactions expressed as (1) are expected, which may be called as a "disproportionation".

Then, in my opinion, the effect of vacuum would be effective for the process (6), and to avoid (3), but you have to check both threshold concentration(4) and threshold temperature(5) to keep the system under 'safe condition' during operation. (Too quick removal of M may lay the solution over the critical concentration of T instantaneously.)

The "decomposition" would be dependent on the temperature and controllable, but the "DECOMPOSITION" tends to take place passing through the thresholds described above and would be uncontrollable once it starts.

The above is my persosal opinion. But remember examples of the critical "peroxide" concentrations of some organic acids over which DECOMPOSITION takes place even at ambient temperatures.

Stefano

#10 Guest_Guest_AndyB_*_*

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Posted 27 August 2006 - 03:49 AM

If you subject the liquid to aerosol generating conditions (atomising nozzles and the like) and generate an aerosol with a MSD of say 0.15 micron would that not have a similar effect in allowing an effect of cracking at lower temperatures due to the vastly increased surface area to volume ratio?
I have seen this with some engine turbine oils.

Andy B




QUOTE (Guest @ Aug 27 2006, 03:01 AM) <{POST_SNAPBACK}>
QUOTE (Shanker Nair @ Sep 21 2004, 12:04 PM) <{POST_SNAPBACK}>

Oh well, thankyou, SO MUCH for your time. I have a clear picture now ! THANKYOU !

Anyway, I just wanetd to elaborate ! What we have is a 40% solution (consisting of trimer(solute) and diethyl malonate(DEM)(solvent)) that decomposes at 250 C. DEM has a BP of 198 C. Our interest is to decompose the trimer and thus obtain the desired product (btw, without adding DEM, the trimer is a crystallien mass and decomposes at 130 C, (and is highly highly explosive).

So, here what we need is DECOMPOSITION, we are not trying to avoid decomposition. My colleagues (seniors) in the lab feel applying vacuum would help to decompose this substance somewhere at abt 150 C. So that DEM is separated at 80 C(from Perry's under vacuum) , and also the cracked/decomposed product is obtained!

Anyway, thans once again :)

Shanker Nair:
The above context quoted meminds me of the following:

(1) The monomer of DEM (also dimer ) and trimer are in an equilibrium state
(M, D, T are monomer,dimer and trimer respectively):
M <--> D <--> T
(2) The "decomposition" of trimer means the reaction:
T--> D -->M in the liquid phase
(3) But the "DECOMPOSITION" of trimer in the solid(or crystalline) or vapor phase means the reaction:
T--> CO2, H2O, ...etc.
(4) There would be a threshold concentration of trimer in the liquid phase for "DECOMPOSITION" with respect to each temperature of trimer solution. (Below 130℃, the trimer will not "DECOMPOSE")
(5) There would be a threshold temperature of solution in the liquid phase for "DECOMPOSITION" with respect to each concentration of trimer (The trimer crystal, 100% trimer, DECOMPOSEs at 130 ℃)
(6) And, you are going to get only 'M' thru evaproation (or distillation) under vacuum from the mixture of M, D, and T. During the process, reactions expressed as (1) are expected, which may be called as a "disproportionation".

Then, in my opinion, the effect of vacuum would be effective for the process (6), and to avoid (3), but you have to check both threshold concentration(4) and threshold temperature(5) to keep the system under 'safe condition' during operation. (Too quick removal of M may lay the solution over the critical concentration of T instantaneously.)

The "decomposition" would be dependent on the temperature and controllable, but the "DECOMPOSITION" tends to take place passing through the thresholds described above and would be uncontrollable once it starts.

The above is my persosal opinion. But remember examples of the critical "peroxide" concentrations of some organic acids over which DECOMPOSITION takes place even at ambient temperatures.

Stefano


#11 sgkim

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Posted 29 August 2006 - 09:27 AM

[quote name='Guest_AndyB_*' date='Aug 27 2006, 03:49 AM' post='7653']
If you subject the liquid to aerosol generating conditions (atomising nozzles and the like) and generate an aerosol with a MSD of say 0.15 micron would that not have a similar effect in allowing an effect of cracking at lower temperatures due to the vastly increased surface area to volume ratio?
I have seen this with some engine turbine oils.

Andy B


Increased surface area will surely increase the reaction of 'decomposition' and diffusion rate.
If M,D, and T are in an equilibrium state, the rate of removal of M will increase the concentration of M and T, therefore the rate of 'decompositon(to M via D)' of T will be increased. So removal of M under vacuum will promote the rate of decomposition of T.

But I expect the "threshold" temperature of DECOMPOSITION(to DO2,..) remain unchanged even with the increased surface.

Stefano




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